# Thread: Error setting up integral for volume in cylindrical coordinates

1. ## Error setting up integral for volume in cylindrical coordinates

Hey, all. Here is the initial problem:

What it the volume of the solid inside both the regions:

r=a\cos\theta

z^2=a^2-r^2

I set the integral up as follows:

\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{a\cos\thet a}2\int_{0}^{\sqrt{a^2-r^2}}r,dz,dr,d\theta

But that doesn't work for some reason, because taking θ from -π/2 to π/2 cancels out, and instead, we have to take θ from π to 2π, like so:

\int_{\pi}^{2\pi}\int_{0}^{a\cos\theta}2\int_{0}^{ \sqrt{a^2-r^2}}r,dz,dr,d\theta

I don't understand why that is.

EDIT: I also don't understand why my LaTeX notation isn't working. Not even simple **** like $r=a$.

EDIT #2: Is it perhaps because both r and θ need to be positive when finding a volume? If so, why would that be true for volume but not area?

2. You appear to be avoiding symmetries. Exploit them, instead.

I didn't look at it real closely, but are you SURE you need [-pi/2,pi/2] or will twice the result from [0,pi/2] do the trick?

3. Originally Posted by TKHunny
You appear to be avoiding symmetries. Exploit them, instead.

I didn't look at it real closely, but are you SURE you need [-pi/2,pi/2] or will twice the result from [0,pi/2] do the trick?
It is symmetrical, and twice the result of [0,pi/2] will indeed do the trick. I also get the correct answer if I take the result of [-pi/2,0] plus that of [0,pi/2], but when I try to simply take the result of [-pi/2,pi/2], some of the terms cancel out, and I get an incorrect answer.

The problem isn't finding the answer, but knowing why one way works and the other doesn't.

The final answer, by the way, is:

[(a^3)(2/9)][3π+4]

4. Well, if things cancel out, this suggests you are crossing an important boundary, like f(x) = x over [-1,1]. What happens at x = 0?

There are a couple of things that seem to be overlooked.

(a^2)^(3/2) = |a| <> a